Solution:

\(\lim_{n\rightarrow\infty}\frac{2n^2+2}{5n^3-3n-1}=\lim_{n\rightarrow\infty}\frac{n^2(2+\frac{2}{n^2})}{n^3(5-\frac{3}{n^2}-\frac{1}{n^3})}\)

\(=\lim_{n\rightarrow\infty}\frac{(2+\frac{2}{n^2})}{n(5-\frac{3}{n^2}-\frac{1}{n^3})}\)

\(=\lim_{n\rightarrow\infty}\frac{1}{n}\times\lim_{n\rightarrow\infty}\frac{(2+\frac{2}{n^2})}{(5-\frac{3}{n^2}-\frac{1}{n^3})}\)

\(=\lim_{n\rightarrow\infty}\frac{1}{n}\times\frac{(2+0)}{(5-0-0)}\)

\(=\lim_{n\rightarrow\infty}\frac{1}{n}\times\frac{2}{5}\)

\(=0\times\frac{2}{5}\)

\(=0\)

\(\lim_{n\rightarrow\infty}\frac{2n^2+2}{5n^3-3n-1}=\lim_{n\rightarrow\infty}\frac{n^2(2+\frac{2}{n^2})}{n^3(5-\frac{3}{n^2}-\frac{1}{n^3})}\)

\(=\lim_{n\rightarrow\infty}\frac{(2+\frac{2}{n^2})}{n(5-\frac{3}{n^2}-\frac{1}{n^3})}\)

\(=\lim_{n\rightarrow\infty}\frac{1}{n}\times\lim_{n\rightarrow\infty}\frac{(2+\frac{2}{n^2})}{(5-\frac{3}{n^2}-\frac{1}{n^3})}\)

\(=\lim_{n\rightarrow\infty}\frac{1}{n}\times\frac{(2+0)}{(5-0-0)}\)

\(=\lim_{n\rightarrow\infty}\frac{1}{n}\times\frac{2}{5}\)

\(=0\times\frac{2}{5}\)

\(=0\)